Polynomials

Board: CBSE | Class: Class 9

Comprehensive study notes for Polynomials by Ajay Yadav (Math King of Katargam). Learn about algebraic expressions, degrees of polynomials, factorization, and algebraic identities.

1. What is a Polynomial?

An algebraic expression of the form P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where aₙ ≠ 0 and n is a whole number, is called a polynomial in variable x.

Examples: 3x² + 2x – 1, 5y³ – 2y + 7, 4z + 1

Non-examples: √x + 2 (exponent is 1/2), 1/x + 3 (exponent is -1)

2. Degree of a Polynomial

The highest power of the variable in a polynomial is called its degree.

Polynomial	Degree	Name
5	0	Constant
2x + 1	1	Linear
3x² – 2x + 1	2	Quadratic
x³ – 3x² + 2x – 1	3	Cubic
x⁴ + 2x³ – x² + 3x – 5	4	Quartic (Biquadratic)

3. Types of Polynomials (by Number of Terms)

• Monomial: One term (e.g., 3x², 5, -2xy)
• Binomial: Two terms (e.g., 2x + 3, x² – 5)
• Trinomial: Three terms (e.g., x² + 2x – 1)

4. Value and Zero of a Polynomial

Value of a Polynomial: P(a) is the value obtained by substituting x = a in P(x).

Zero/Root of a Polynomial: A number a such that P(a) = 0 is called a zero of the polynomial.

Example: For P(x) = x² – 5x + 6, P(2) = 4 – 10 + 6 = 0, so x = 2 is a zero.

5. Remainder Theorem

If a polynomial P(x) is divided by (x – a), the remainder is P(a).

Proof: P(x) = (x – a)Q(x) + R, where R is a constant (since divisor is of degree 1). Substituting x = a gives P(a) = R.

Example: Find the remainder when x³ – 3x² + 2x – 1 is divided by (x – 2).

P(2) = 8 – 12 + 4 – 1 = -1

6. Factor Theorem

(x – a) is a factor of P(x) if and only if P(a) = 0.

Example: Check if (x – 3) is a factor of x³ – 6x² + 11x – 6.

P(3) = 27 – 54 + 33 – 6 = 0. Yes, (x – 3) is a factor.

7. Algebraic Identities

Memorize these key identities:

• (a + b)² = a² + 2ab + b²
• (a – b)² = a² – 2ab + b²
• a² – b² = (a + b)(a – b)
• (x + a)(x + b) = x² + (a + b)x + ab
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a – b)³ = a³ – 3a²b + 3ab² – b³
• a³ + b³ = (a + b)(a² – ab + b²)
• a³ – b³ = (a – b)(a² + ab + b²)

8. Factorization Techniques

Method 1: Common Factor

Example: 6x²y + 9xy² = 3xy(2x + 3y)

Method 2: Grouping

Example: x³ + x² + x + 1 = x²(x + 1) + 1(x + 1) = (x + 1)(x² + 1)

Method 3: Splitting the Middle Term

Example: Factorise x² + 7x + 12. Find p, q such that p + q = 7 and pq = 12. p = 3, q = 4.

= x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)

Solved Examples

Example 1: Using identities, evaluate 102².

Solution: (100 + 2)² = 100² + 2(100)(2) + 2² = 10000 + 400 + 4 = 10404

Example 2: Factorise x³ – 8.

Solution: Using a³ – b³ = (a – b)(a² + ab + b²), where a = x, b = 2:

= (x – 2)(x² + 2x + 4)

Example 3: Find the value of k if (x – 1) is a factor of 4x³ + 3x² – 4x + k.

Solution: P(1) = 4(1)³ + 3(1)² – 4(1) + k = 4 + 3 – 4 + k = 3 + k

For (x – 1) to be a factor, P(1) = 0, so 3 + k = 0 ⟹ k = -3

Practice Questions

1. Find the degree of the polynomial 5x³ – 2x² + 7x – 1.
2. Find the zero of the polynomial P(x) = 2x + 5.
3. Factorise: x² + 9x + 18
4. Factorise: x³ – 3x² – 9x – 5
5. Evaluate using identities: 104 × 96
6. If x + 1/x = 5, find x² + 1/x².
7. Expand: (a – b + c)²

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